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Cone area

  1. Sep 30, 2005 #1
    This is a problem I had in a test and almost everyone got different answers for it, we discussed and well, I spotted mistakes in their solutions so I think mine is right but I wanted to check here and also ask if there is an easier/faster way to do it.

    There is a container that is similar to the bottom part of a cone which is cut into half. Top radius is [tex]10cm[/tex] and bottom is [tex]20cm[/tex]. And the volume of this container is [tex]500cm^{3}[/tex]. What is the length of the slanted side ?

    So what I did was, first I drew this. ;P (See attachment)

    Then what I did was to write an equation for [tex]V_2[/tex] in terms of [tex]V_1[/tex] and [tex]V_T[/tex]

    By solving the equation for height, I got [tex]h=1.5915[/tex]
    Then I used Pythagoras' theorem to find the length of the slanted side and it comes to:
    [tex]10^2+2h^2=s^2=>100+3.183^2=s^2=> s=10.4944[/tex]

    Is this correct ? And is there a formula to find the volume of this shape ?
     

    Attached Files:

  2. jcsd
  3. Sep 30, 2005 #2

    Tide

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    The volume of the "flower pot" section is

    [tex]V = \frac {\pi h}{3} (R^2 + rR + r^2)[/tex]

    where R is the large radius, r is the small radius and h is the same as your x. It's simply the difference between the volume of the large cone and the small cone.
     
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